for any g,h in φ(G), we have φ(x) = g and φ(y) = h, for SOME x,y in G (because φ is onto φ(G)). but we know that ,

so . but:

, which quickly gives:

the quantity is called the commutator of x and y, and is written [x,y]. and here's a sneak peek of what lies ahead:

the subgroup of G generated by all the commutators, is called the commutator subgroup [G,G] (or sometimes G'). and what you (we?) have just proved is that
G/[G,G] is an abelian group (think about this for a second until it dawns on you what i am saying is true).

well, we've almost proved that. we haven't yet shown that [G,G] is normal in G, but it is, and if φ:G-->H is a homomorphism, and H is abelian, then ker(φ) contains [G,G]. [G,G] measures how "non-abelian" G is, in a natural way (the center of G,Z(G), does the same thing, but in a way that is "not natural").